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Lets $S=\{z \in C:|z-1|=1$ and $(\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2}\}$. Let $\mathrm{z}_1, \mathrm{z}_2$ $\in S$ be such that $\left|z_1\right|=\max _{z \in S}|z|$ and $\left|z_2\right|=\min _{z \in S}|z|$. Then $\left|\sqrt{2} z_1-z_2\right|^2$ equals :
$1$
$4$
$3$
$2$
Solution
Let $Z=x+i y$
Then $(x-1)^2+y^2=1 \rightarrow(1)$
$(\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2}$
$\quad \Rightarrow(\sqrt{2}-1) x+y=\sqrt{2} \rightarrow(2)$
Solving $(1) and (2)$ we get
Either $\mathrm{x}=1$ or $x=\frac{1}{2-\sqrt{2}} \rightarrow$ $(3)$
On solving $(3)$ with $(2)$ we get
For $x=1 \Rightarrow y=1 \Rightarrow Z_2=1+i$ and for
$x=\frac{1}{2-\sqrt{2}} \Rightarrow y=\sqrt{2}-\frac{1}{\sqrt{2}} \Rightarrow Z_1=\left(1+\frac{1}{\sqrt{2}}\right)+\frac{i}{\sqrt{2}}$
Now
$\left|\sqrt{2} z_1-z_2\right|^2$
$=\left|\left(\frac{1}{\sqrt{2}}+1\right) \sqrt{2}+i-(1+i)\right|^2$
$=(\sqrt{2})^2$
$=2$
Similar Questions
Let $z_k=\cos \left(\frac{2 k \pi}{10}\right)+ i \sin \left(\frac{2 k \pi}{10}\right) ; k =1,2, \ldots 9$.
| List $I$ | List $II$ |
| $P.$ For each $z_k$ there exists a $z_j$ such that $z_k \cdot z_j=1$ | $1.$ True |
| $Q.$ There exists a $k \in\{1,2, \ldots ., 9\}$ such that $z_{1 .} . z=z_k$ has no solution $z$ in the set of complex numbers. | $2.$ False |
| $R.$ $\frac{\left|1-z_1\right|\left|1-z_2\right| \ldots . .\left|1-z_9\right|}{10}$ equals | $3.$ $1$ |
| $S.$ $1-\sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)$ equals | $4.$ $2$ |
Codes: $ \quad P \quad Q \quad R \quad S$
